To write the expression of four-force in STR \(\ F=\frac{dp}{dt}=\ m\gamma\ a +\ m\gamma\frac{\ u .\ a}{\ c^2 -\ u^2}\ u\) Here a is acceleration I used the equation \(\ p=\gamma\ m\ u\) I interpreted F as four force,p as four momentum, a as four-acceleration, u as four velocity etc... Mere differentitation is giving the answer;But I do not know if the method is correct.Because, \(\ p=\gamma\ m\ u\) for 3 velocity---that's for sure.But is it also true for four velocity and four momentum?
It looks to me expression of force is not intended to mean four-force...this is the expression for 3 force.
Hi, The four-velocity and four-acceleration are defined as : \( \begin{align} v^{\mu} \,&=\, \frac{ \text{d} x^{\mu} }{ \text{d} \tau } \\ a^{\mu} \,&=\, \frac{ \text{d}^{2} x^{\mu} }{ \text{d} \tau^{2} } \end{align} \)Where \(\tau\) is the proper time. On the other hand, the classical three-velocity is defined as a derivative with respect to \(t\) : \( \bar{u} \,=\, \frac{\text{d}\bar{x}}{\text{d}t} \)Since \(\text{d}\tau \,=\, \frac{ \text{d}t }{ \gamma }\), the spatial components of the four-velocity are related to the components of the three-velocity by \(v^{i} \,=\, \gamma \, u_{i}\) for \(i \,=\, x,\, y,\, z\), so basically the extra factors of \(\gamma\) that you seem to be worrying about are absorbed into the definition of the four-velocity: \(v^{\mu} \,=\, ( \gamma c, \, \gamma \bar{u} )\). The four-momentum is defined as : \( p^{\mu} \,=\, m v^{\mu} \)Extracting the spatial components on both sides gets you the correct three-vector relation: \(\bar{p} \,=\, \gamma m \bar{u}\). Finally, Newton's second law in covariant form is : \(f^{\mu} \,=\, \frac{\text{d} p^{\mu}}{\text{d} \tau}\)or simply : \(f^{\mu} \,=\, m a^{\mu}\) Does that clarify things?
After posting this thread,I checked once more and got the result you have given. The expression of 4-force is not as is given in this expression.This is applicable to three force.
Are you sure there even is a covariant expression for Newton's second law? Also, even if you're just looking at the spacial portion of the vector, shouldn't it read \(\vec{F}=\frac{d\vec{p}}{dt}\) instead of \(\vec{F}=m\frac{d\vec{p}}{d\tau}\)?
Yes (Wiki's article on it is here) though I don't think there's really much use for it. For empirical laws like friction and Hooke's law we know they're not fundamental forces so we don't really care about expressing them in covariant form. On the other hand, gravity isn't really a force, and two of the fundamental forces (weak and strong) are only effective at subatomic scales where QFTs take over and Lagrangians and Hamiltonians reign supreme. I've only seen a four-force given for classical electrodynamics - the Lorentz force in covariant form is: \(f^{\mu} \,=\, q F^{\mu \nu} v_{\nu}\)where \(F^{\mu \nu}\) is the field strength tensor. Extracting the spatial components will get you the more familiar \(\bar{f} \,=\, q \bigl[ \bar{E} \,+\, \bar{v} \,\times\, \bar{B} \bigr]\). Yup - I've just corrected my post accordingly.
Thanks for the link. If I understand it correctly, it's similar to an idea I was thinking of when I first read this post. The thing is, as far as I can tell, the contravariant 4-vector defined in that article isn't the actual force on the object. The spatial part of it reads "\(\gamma\vec{\mathbf{f}}\)", so you need to divide by gamma to get the actual physical force. That's why I was wondering if your time derivative should be \(\frac{dp}{dt}\) instead of \(\frac{dp}{d\tau}\), in addition to the extra mass factor typo.
Well in relativistic mechanics \(f^{\mu} \,=\, m a^{\mu}\) defines the four-force the same way \(\bar{F} \,=\, m \bar{a}\) defines the three-force in non-relativistic mechanics. If one of the two should be considered more "physical" than the other I'd go with the four-force - you get the regular three-force back when \(\gamma \approx 1\), which is either asymptotically in the limit \(v \rightarrow 0\) (ie. non relativistic conditions) or \(c \rightarrow \infty\) (Galilean relativity). Similar relationships hold between the relativistic and non-relativistic kinetic energy and momentum for example. Also with the four-vectors the reason proper time derivatives appear is that the proper time is a Lorentz scalar. Part of the definition of a four-vector is that its transformation law should be \(v^{\mu} \,\mapsto\, \Lambda^{\mu}_{\text{ }\nu} v^{\nu}\) under a Lorentz transformation, and a scalar derivative of a four-vector is still a four-vector. So you have: \( \frac{\text{d}}{\text{d}\tau} \, \bigl( \Lambda^{\mu}_{\text{ }\nu} \, x^{\nu} \bigr) \:=\: \Lambda^{\mu}_{\text{ }\nu} \, \frac{\text{d}x^{\nu}}{\text{d}\tau} \)and \(\frac{\text{d}x^{\mu}}{\text{d}\tau}\) is a genuine four-vector, but in general \( \frac{\text{d}}{\text{d}t^{\prime}} \, \bigl( \Lambda^{\mu}_{\text{ }\nu} \, x^{\nu} \bigr) \:\neq\: \Lambda^{\mu}_{\text{ }\nu} \, \frac{\text{d}x^{\nu}}{\text{d}t} \)simply because \(t^{\prime} \,\neq\, t\), and you'll usually only see time derivatives appear as part of a four-gradient or four-divergence.
Yeah, I've been through the deductions before and I know why \(d\tau\) is used. Just wondered if there was a way of writing the force law covariantly in a more direct manner, so that the spatial part of the 4-force actually represents the relativistic force rather than this force times a factor of gamma. I think this is probably the most direct covariant expression we can get though.
Heh, I've probably "helpfully" informed posters of what they already know dozens of times by now. It's one of the limitations of a message board - I can't read body language or minds like I normally would otherwise Please Register or Log in to view the hidden image! Same here, as far as I know. dp[sub]i[/sub] / dt just doesn't have the right transformation law to be packed directly into a four-vector.
